Python 如何筛选数据框中的行,其中列的值等于列表的某个值
我有一个数据框,有两列:一列表示ID\u编号,一列表示week\u编号。 它可以是这样的:Python 如何筛选数据框中的行,其中列的值等于列表的某个值,python,pandas,dataframe,pandas-groupby,Python,Pandas,Dataframe,Pandas Groupby,我有一个数据框,有两列:一列表示ID\u编号,一列表示week\u编号。 它可以是这样的: df1 = pd.DataFrame({'ID_number':[13, 13, 14, 14, 14, 15, 15,16], 'week_number':[1, 2, 1, 2, 3, 1, 4, 5]}) # ID_number week_number #0 13 1 #1 13 2 #2 14 1 #3 14 2 #4 14 3 #5 15 1 #6 15 4
df1 = pd.DataFrame({'ID_number':[13, 13, 14, 14, 14, 15, 15,16], 'week_number':[1, 2, 1, 2, 3, 1, 4, 5]})
# ID_number week_number
#0 13 1
#1 13 2
#2 14 1
#3 14 2
#4 14 3
#5 15 1
#6 15 4
#7 16 5
我想为每个不同的ID选择,那些周值为2和3的ID,然后为数据创建一个标签。如果一个ID没有第2周和第3周,我会将其标记为1。否则,我会将其标记为0
目前,我提出了一个相当不优雅的解决方案,这是可行的,但我相信一定有另一种方法:
def check_courier_week(df, field, weeks):
weeks_not_provided = weeks
new_df = df
new_df['label'] = np.zeros(len(df))
for c in np.unique(df[field]):
tmp = df[df[field] == c]
if len(np.unique(tmp.week_number.isin(weeks_not_provided))) == 1 and np.unique(np.unique(tmp.week_number.isin(weeks_not_provided))) == False:
new_df['label'][df[field] == c] = 1
else:
new_df['label'][df[field] == c] = 0
return new_df
有没有关于如何改进的想法?我想可能有一个使用groupby的解决方案,但我不知道如何实现它
结果标签应该是:
# ID_number week_number label
#0 13 1 0.0
#1 13 2 0.0
#2 14 1 0.0
#3 14 2 0.0
#4 14 3 0.0
#5 15 1 1.0
#6 15 4 1.0
#7 16 5 1.0
谢谢 使用
groupby
和transform
any
(~(df1['week_number'].isin([2,3])).groupby(df1['ID_number']).transform('any')).astype(int)
Out[39]:
0 0
1 0
2 0
3 0
4 0
5 1
6 1
7 1
Name: week_number, dtype: int32
回答如何使用groupby:您可以按ID_编号进行分组,然后以这种方式找到标签,即:
df1['label'] = np.zeros(len(df))
grouped_table = df1.groupby('ID_number')
groups = list(set(df1['ID_number']))
for group in groups:
test_list = list(set(grouped_table.getgroup(group)))
if (2 in test_list) & (3 in test_list):
df1.loc[df1['ID_number'] == group]['label'] = 0
else:
df1.loc[df1['ID_number'] == group]['label'] = 1
使用和不使用分组:
unique = df1.loc[df1['week_number'].isin([2,3]), 'ID_number'].unique()
df['label'] = np.where(df1['ID_number'].isin(unique), 0, 1)
或:
虽然效率不高,但您可以通过以下方式利用
set
操作:
使用groupby,然后可以使用每组周数的最小值和最大值来确定标签的值
df['label'] = (~df1['ID_number'].isin(unique)).astype(int)
print(df)
ID_number week_number label
0 13 1 0
1 13 2 0
2 14 1 0
3 14 2 0
4 14 3 0
5 15 1 1
6 15 4 1
7 16 5 1
def checker(x):
return set(x).isdisjoint({2, 3})
df1['flag'] = df1.groupby('ID_number')['week_number'].transform(checker)
print(df1)
ID_number week_number flag
0 13 1 0
1 13 2 0
2 14 1 0
3 14 2 0
4 14 3 0
5 15 1 1
6 15 4 1
7 16 5 1